Question

A particle released from top of a tower reaches ground after 12 sec calcute the following 1.Height of tower 2.distance travelled by tower in 12th sec
3 . Distance travelled in 10 sec
4. Distance travelled in 10th sec
5. distance travelled in 5th , 7th, and 9th sec ​

Answer 1

Answer:

The formula here will be used as.

s= ut +1/2 at square

intial speed of stone = 0m/s

time taken by the particle =12 sec

height of the tower..

H=ut +1/2 at square

H=0x12+1/2x10x12x12

......Gravitation Constant = 10....

H=0+1/2x144

H=72m

1...height of tower is 72 m

2..distance in 12 sec = 72 m

3..again use the formula

instead of 12 take 10....

H=0+1/2x100

H=50 m

For 10 sec..50 m

4.....50 m..

5 ......In 5 sec...

H=0+1/2x25

H=12.5 m

....In 7 sec...

H=0+1/2x49

H=24.5 m

........In 9 sec....

H=0+1/2x81

H=40.5 m..